geometry-sample-test-items.pdf

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SAMPLE TEST ITEMS

October 2013

Geometry

This document, originally published in 2013,contains information relating to a transition to PARCC testing;

however, at this time, there is no plan to transition to PARCC. The sample items and student work reflect the

current EOC Geometry assessment; therefore, teachers are encouraged to use the samples provided in this

document as additional resources, but should uset

he current Assessment Guidance Geometry document for

up-to-date EOC testing information.


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Louisiana State Board of Elementary and Secondary Education

Mr. Charles E. Roemer

President

Sixth BESE District

Ms. Kira Orange Jones

Second BESE District

Ms. Carolyn Hill

Eighth BESE District

Mr. James D. Garvey, Jr.Ms. Lottie P. Beebe

Third BESE DistrictMs. Connie BradfordMember-at-Large

Vice President

First BESE District Mr. Walter Lee Dr. Judy Miranti

Fourth BESE District Member-at-Large

Ms. Holly Boffy

Secretary/Treasurer Mr. Jay Guillot Mr. Stephen Waguespack

Seventh BESE District Fifth BESE District Member-at-Large

Ms. Heather Cope

Executive Director

The mission of the Louisiana Department of Education (LDOE) is to ensure equal access to education and to promoteequal excellence throughout the state. The LDOE is committed to providing Equal Employment Opportunities and iscommitted to ensuring that all its programs and facilities are accessible to all members of the public. The LDOE doesnot discriminate on the basis of age, color, disability, national origin, race, religion, sex, or genetic information. Inquiriesconcerning the LDOEs compliance with Title IX and other civil rights laws may be directed to the Attorney, LDOE, Officeof the General Counsel, P.O. Box 94064, Baton Rouge, LA 70804-9064; 877-453-2721 or [email protected] about the federal civil rights laws that apply to the LDOE and other educational institutions is available onthe website for the Office of Civil Rights, USDOE, athttp://www.ed.gov/about/offices/list/ocr/.

This project is made possible through a grant awarded by the State Board of Elementary and Secondary Educationfrom the Louisiana Quality Education Support Fund8(g).

This public document was published at a total cost of $6,000.00. This web-only document was published forthe Louisiana Department of Education, Office of Assessments, P.O. Box 94064, Baton Rouge, LA 70804-9064,by Pacific Metrics Corporation, 1 Lower Ragsdale Drive, Building 1, Suite 150, Monterey, CA 93940. This materialwas published in accordance with the standards for printing by state agencies established pursuant to R.S. 43:31and in accordance with the provisions of Title 43 of the Louisiana Revised Statutes.

For further information, contact: Louisiana

Department of Educations Help Desk

1-877-453-2721Ask LDOE?

https://www.louisianabelieves.com/resources/ask-ldoe

2013 by Louisiana Department of Education
mailto:[email protected]://www.ed.gov/about/offices/list/ocr/http://www.ed.gov/about/offices/list/ocr/https://www.louisianabelieves.com/resources/ask-ldoehttps://www.louisianabelieves.com/resources/ask-ldoehttps://www.louisianabelieves.com/resources/ask-ldoehttps://www.louisianabelieves.com/resources/ask-ldoehttps://www.louisianabelieves.com/resources/ask-ldoehttps://www.louisianabelieves.com/resources/ask-ldoehttp://www.ed.gov/about/offices/list/ocr/mailto:[email protected]


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Table of Contents

Introduction .......................................................................................... 1

Purpose of This Document ....................................................................... 1

Geometry Administration ......................................................................... 1

EOC Achievement Levels ......................................................................... 2

EOC Achievement-Level Definitions ......................................................... 2

Geometry .............................................................................................. 3

Constructed-Response Rubric ................................................................. 3

Testing Materials and Online Tools .......................................................... 4

Multiple-Choice Items ........................................................................... 5

Constructed-Response Item ................................................................. 31

Constructed-Response Item ................................................................... 31

Scoring Information ............................................................................... 34

Sample Student Responses .................................................................... 35

Geometry Typing Help ........................................................................... 44

Geometry Reference Sheet ..................................................................... 46


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1Geometry EOC Sample Test ItemsOctober 2013

Introduction

Louisiana Believes embraces the principle that all children can achieve at highlevels, as evidenced in Louisianas recent adoption of the Common Core StateStandards (CCSS). Louisiana Believes also promotes the idea that Louisianas

educators should be empowered to make decisions to support the successof their students. In keeping with these values, the Department has createddocuments with sample test items to help prepare teachers and students asthey transition to the CCSS. The documents reflect the States commitment toconsistent and rigorous assessments and provide educators and families withclear information about expectations for student performance.

Purpose of This Document

Teachers are encouraged to use the sample items presented in this documentin a variety of ways to gauge student learning and to guide instruction anddevelopment of classroom assessments and tasks. The document includesmultiple-choice and constructed-response items that exemplify how theCommon Core State Standards for Mathematics (CCSSM) will be assessedon the End-of-Course (EOC) tests. A discussion of each item highlights theknowledge and skills the item is intended to measure.

As Louisiana students and teachers transition to the CCSS and the Partnershipfor Assessment of Readiness for College and Careers (PARCC) assessments, theGeometry EOC assessment will include only items aligned to the CCSS.

As you review the items, it is important to remember that the sample itemsincluded in this document represent only a portion of the body of knowledge

and skills measured by the EOC test.

Geometry Administration

The Geometry EOC test is administered to students who have completed oneof the following courses:

Geometry: course code 160323

Applied Geometry: course code 160332


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EOC Achievement Levels

Student scores for the Geometry EOC test are reported at four achievementlevels: Excellent, Good, Fair, and Needs Improvement. General definitions of theEOC achievement levels are shown below.

EOC Achievement-Level Definitions

Excellent:A student at this achievement level has demonstrated masteryof course content beyond Good.

Good:A student at this achievement level has demonstrated mastery ofcourse content and is well prepared for the next level of coursework in thesubject area.

Fair:A student at this achievement level has demonstrated only thefundamental knowledge and skills needed for the next level of courseworkin the subject area.

Needs Improvement:A student at this achievement level has notdemonstrated the fundamental knowledge and skills needed for the next levelof coursework in the subject area.

Because of the shift from grade-level expectations to the CCSS, this documentdiffers from the Released Test Item Documents. Many of the released items frompast test administrations may not be indicative of the types of items on theupcoming December and May EOC transitional assessments. To better alignthe transitional test to the content of the CCSS, new items were developed.Therefore, this document includes sample items, rather than releaseditems. These sample items reflect the way the CCSS will be assessed and

represent the new items that students will encounter on the transitional EOCassessments. Because these are not released items, item-specific informationabout achievement levels is not included.


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Geometry

The Geometry EOC test contains forty-six multiple-choice items and oneconstructed-response item. In addition, some field test items are embedded.

Multiple-choice items assess knowledge, conceptual understanding, and

application of skills. They consist of an interrogatory stem followed by fouranswer options and are scored as correct or incorrect.

Constructed-response items require students to compose an answer, and theseitems generally require higher-order thinking. A typical constructed-responseitem may require students to develop an idea, demonstrate a problem-solvingstrategy, or justify an answer based on reasoning or evidence. The Geometryconstructed-response item is scored on a scale of 0 to 4 points. The generalconstructed-response rubric, shown below, provides descriptors for eachscore point.

Constructed-Response Rubric

Score Description

4

The students response demonstrates in-depth understanding ofthe relevant content and/or procedures.

The student completes all important components of the taskaccurately and communicates ideas effectively.

Where appropriate, the student offers insightful interpretationsand/or extensions.

Where appropriate, the student uses more sophisticatedreasoning and/or efficient procedures.

3

The student completes the most important aspects of the taskaccurately and communicates clearly.

The students response demonstrates an understanding of majorconcepts and/or processes, although less important ideas ordetails may be overlooked or misunderstood.

The students logic and reasoning may contain minor flaws.

2

The student completes some parts of the task successfully. The students response demonstrates gaps in conceptual

understanding.

1 The student completes only a small portion of the task and/or

shows minimal understanding of the concepts and/or processes.

0 The students response is incorrect, irrelevant, too brief to

evaluate, or blank.


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Testing Materials and Online Tools

Students taking the Geometry EOC test have access to a number of resources.Scratch paper, graph paper, and pencils are provided by test administrators andcan be used by students during all three sessions of the Geometry EOC test.

There are also buttons at the top of the screen that a student may click to openthe online tools. The list below identifies the online tools available for eachsession. Please note that the rulers and/or protractor may not be available forsome items.

Session 1

Geometry Reference Sheet

protractor

inch ruler

centimeter ruler

Note: Students are NOT allowed to use calculators during session 1 unlessthey have the approved accommodation Assistive Technology and are

allowed the use of a calculator.

Sessions 2 and 3

Geometry Reference Sheet

calculator

protractor

inch ruler

centimeter ruler

Also available in session 2, which contains the constructed-response item,is the Geometry Typing Help (see pages 44 and 45). This online tool describeshow to enter special characters, symbols, and formatting into typed responses.The graph paper, Geometry Reference Sheet, Geometry Typing Help, andEOC Tests online calculator can be found on the EOC Tests homepage atwww.louisianaeoc.orgunder Test Coordinator Materials: Testing Materials.
http://www.louisianaeoc.org/http://www.louisianaeoc.org/


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Multiple-Choice Items

This section presents ten multiple-choice items selected to illustrate the typeof skills and knowledge students need in order to demonstrate understandingof the CCSS in the Geometry course. These items also represent the skills

students need in order to meet performance expectations for Math Practices.Information shown for each item includes

conceptual category,

domain,

cluster,

standard,

Math Practices,

calculator designation (calculator allowed or calculator not allowed),

tool restrictions, correct answer,

commentary on the skills and knowledge associated with the standardmeasured by the item,

commentary on the Math Practices linked with the item,

commentary on why the correct answer is correct including, in some cases,how the answer is achieved, and

commentary on each answer choice, explaining why it is correct orincorrect.


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Conceptual Category: GGeometry

Domain: COCongruence

Cluster: DMake geometric constructions

Standard: 12Make formal geometric constructions with avariety of tools and methods (compass and

straightedge, string, reflective devices, paperfolding, dynamic geometric software, etc.).Copying a segment; copying an angle; bisecting

a segment; bisecting an angle; constructing

perpendicular lines, including the perpendicular

bisector of a line segment; and constructing aline parallel to a given line through a point not on

the line.

Math Practices: 5. Use appropriate tools strategically.6. Attend to precision.

7. Look for and make use of structure.Calculator: Calculator allowed

Tool Restrictions: None


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Melanie wants to construct the perpendicular bisector of line segment ABusing a compass and straightedge.

A B

Which diagram shows the first step(s) of the construction?

A. A B

B. A B

C. A B

D. A B

*correct answer


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This item requires students to identify the initial steps of constructing aperpendicular bisector. Constructions are typically difficult to assess in amultiple-choice format. However, asking students to analyze which figure willresult in the given construction assesses reasoning about constructions withoutthe need for providing geometric tools during testing or having to manually scoreeach response.

This item is linked to three of the Math Practices. Math Practice 5 (Use appropriate tools strategically.): Students must recognize

how to effectively use a compass to begin constructing a perpendicularbisector of a line segment.

Math Practice 6 (Attend to precision.): Students must recognize that onlyone method shown guarantees a precise construction.

Math Practice 7 (Look for and make use of structure.): Students must usethe structure of the two congruent circles and recognize that the circlesintersect at points that are equidistant from the endpoints of the givenline segment. These points may be connected to construct a perpendicularbisector.

Option D: This is the correct answer. The figure shows two circles that appearto be congruent centered at the endpoints of the given line segment.Connecting the intersection points of these two circles guarantees the bisectingline created is perpendicular (at a 90 degree angle) to segment AB.

Option A: The student incorrectly draws two circles that appear to be congruentand intersect at what appears to be the midpoint of segment AB; however, thefigure does not show how a perpendicular bisector could be constructed, as itdoes not guarantee that a line drawn would intersect at a 90 degree angle.

Option B: The student incorrectly draws one circle that appears to be centeredat the midpoint of segment AB. The midpoint is where a bisector will intersect,but this figure does not show how to ensure the bisector will be perpendicular.

Option C: The student incorrectly draws two circles with different radii centeredat the endpoints of the given line segment. Connecting the intersection pointsof these two circles will result in a line perpendicular to segment AB that doesnot bisect it.


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Domain: COCongruence

Cluster: CProve geometric theorems

Standard: 11Prove theorems about parallelograms.Theorems include: opposite sides are congruent,

opposite angles are congruent, the diagonals of aparallelogram bisect each other, and conversely,

rectangles are parallelograms with congruent

diagonals.

Math Practices: 1. Make sense of problems and persevere insolving them.

3. Construct viable arguments and critique thereasoning of others.

Calculator: Calculator allowed



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Use the figure to answer the question.

B C

A D

Missy is proving the theorem that states that opposite sides of aparallelogram are congruent.

Given: Quadrilateral ABCD is a parallelogram.

Prove: AB CD and BC DA

Missys incomplete proof is shown.

Statement Reason

1. Quadrilateral ABCD is aparallelogram.

1. given

2. AB || CD; BC || DA 2. definition of parallelogram

3. ? 3. ?4. AC AC 4. reflexive property

5. ABC CDA 5. angle-side-anglecongruence postulate

6. AB CD and BC DA 6. Corresponding parts ofcongruent triangles arecongruent (CPCTC).

Which statement and reason should Missy insert into the chart as step 3to complete the proof?

A. BD BD; reflexive property

B. AB CD and BC DA ; reflexive property

C. ABD CDB and ADB CBD; When parallel lines are cutby a transversal, alternate interior angles are congruent.

D. BAC DCA and BCA DAC; When parallel lines are cut

by a transversal, alternate interior angles are congruent.

*correct answer


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This item requires students to provide the missing statement and reasonin a two-column proof of the theorem that says that the opposite sides of aparallelogram are congruent. Assessing students abilities to write proofs in amultiple-choice item is difficult. This item allows students to demonstrate theirreasoning skills by asking for one of the middle steps in the proof, includingboth the statement and reason.

This item is linked to three of the Math Practices. Math Practice 1 (Make sense of problems and persevere in solving them.):

Students must analyze the given information in the item and the constraintsof the details provided in the proof, and understand the goals of the problem.

Math Practice 3 (Construct viable arguments and critique the reasoning ofothers.): Students must complete the logical progression presented in theproof.

Option D: This is the correct answer. Students are given a set of correspondingcongruent sides in statement and reason 4. This leads into proving trianglecongruence using the angle-side-angle congruence postulate in statementand reason 5. In order to reach statement and reason 5, students need toprovide two pairs of corresponding congruent angles, which include the givencorresponding congruent sides in statement and reason 4.

Option A: This answer is a true statement with a correct reason for thestatement based on the given information; however, it is not relevant tocompleting the proof because it does not support statement and reason 5.

Option B: The student uses the final result of the proof as the statement andprovides an incorrect reason for the statement.

Option C: This response is also a true statement with a correct reason, but it

is not relevant to the given proof because it does not support statement andreason 5.


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Domain: SRTSimilarity, Right Triangles, and Trigonometry

Cluster: AUnderstand similarity in terms of similaritytransformations

Standard: 1Verify experimentally the properties of dilations

given by a center and a scale factor:a. A dilation takes a line not passing through thecenter of the dilation to a parallel line, and leavesa line passing through the center unchanged.

Math Practices: 7. Look for and make use of structure.

Calculator: Calculator not allowed



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2

2

Rosa graphs the line y = 3x + 5. Then she dilates the line by

a factor of15

with (0, 7) as the center of dilation.

y

10

8

6

10 8 6 4 4 6 8 10 x

4

6

8

10

Which statement best describes the result of the dilation?

1

A.The result is a different line

5the size of the

original line.

B. The result is a different line with a slope of 3.

C. The result is a different line with a slope of 1

.3

D. The result is the same line.

*correct answer

This item requires students to evaluate the effect of a dilation of a line notpassing through the center of the dilation. Specifically, students must recognizethat the dilated line will have the same slope as the original line.

This item is linked to one of the Math Practices.

Math Practice 7 (Look for and make use of structure.): Students mustrecognize that the effect of the described dilation would not change theslope of the line, and that it would not be the same line either.


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Option B: This is the correct answer. The result of a dilation of a line notpassing through the center of the dilation is a line parallel to the original line.Therefore, the slope will remain the same.

Option A: The student confuses the line with line segment. A dilation of a linesegment with a scale factor of 1/5 would result in a smaller line segment;however, a dilation of a line results in a line.

Option C: The student incorrectly assumes that the result of the dilation is aline that is perpendicular to the original line.

Option D: The student incorrectly assumes that the result of the dilation is theoriginal line; however, this is only true of lines that pass through the center ofthe dilation.


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Cluster: AUnderstand similarity in terms of similaritytransformations

Standard: 2Given two figures, use the definition of similarity

in terms of similarity transformations to decideif they are similar; explain using similaritytransformations the meaning of similarity fortriangles as the equality of all correspondingpairs of angles and the proportionality of allcorresponding pairs of sides.






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2

Use the graph to answer the question.

y

10

P B' 8 B

6

10 8

4

A' C'

6 4 4 6 8 10x

4

6

A C8

10

Kelly dilates triangle ABC using point P as the center of dilation and createstriangle A'B'C'.

By comparing the slopes of AC and CB and A'C' and C'B', Kelly found thatACB and A'C'B' are right angles.

Which set of calculations could Kelly use to prove ABC is similar to A'B'C'?

A.slope AB =

7 (7)

=14

= 2

2 (5) 7

7 3 4slope AB=3 (5)

=2

= 2

B. AB2 = 72 + 142

AB2 = 22 + 42

C. AC 7tan ABC = =

BC 14

tan ABC=AC = 2

BC 4

D. ABC+BCA +CAB =180

ABC+BCA +CAB= 180

*correct answer


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This item requires students to decide which evidence is sufficient for provingthat the given triangles are similar. Providing the calculations in the answerchoices eliminates the possibility of a student missing the item due to anarithmetic error. This allows the item to focus on assessing the studentsreasoning in determining the AA criterion for triangle similarity.

This item is linked to two of the Math Practices.

Math Practice 1 (Make sense of problems and persevere in solving them.):Students must analyze the given information in the item and the constraintsof the details provided, and understand the goals of the problem.

Math Practice 3 (Construct viable arguments and critique the reasoningof others.): Students must determine which calculation is sufficient forconstructing a proof showing the given triangles are similar.

Option C: This is the correct answer. The tangents of the correspondingangles ABC and A'B'C' are equal. This shows that the corresponding anglesare congruent and the corresponding sides have the same constant ofproportionality.

Option A: The student incorrectly assumes that congruent correspondingslopes in the triangles imply that corresponding angles are congruent.Using the slopes is a possible method, but additional steps are neededto prove that the triangles are similar.

Option B: The student incorrectly selects a set of calculations that helpdetermine one pair of corresponding side lengths. This is not sufficient forproving triangle similarity.

Option D: The student incorrectly assumes that because all triangles have3 angles whose measures sum to 180 degrees, the triangles are similar.


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Cluster: BProve theorems involving similarity

Standard: 5Use congruence and similarity criteria fortriangles to solve problems and to prove

relationships in geometric figures.





Hector knows two angles in triangle A are congruent to twoangles in triangle B. What else does Hector need to knowto prove that triangles A and B are similar?

A. Hector does not need to know anything else abouttriangles A and B.

B. Hector needs to know the length of anycorresponding side in both triangles.

C. Hector needs to know all three angles in triangleA are congruent to the corresponding angles intriangle B.

D. Hector needs to know the length of the sidebetween the corresponding angles on each triangle.

*correct answer

This item requires students to determine which information is needed tocomplete an argument that two triangles are similar. Specifically, students

must recognize that two pairs of corresponding congruent angles are sufficientfor proving two triangles are similar.


Math Practice 1 (Make sense of problems and persevere in solving them.):Students must analyze the given information in the item and the constraintsof the details provided, and understand the goals of the problem.


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Math Practice 3 (Construct viable arguments and critique the reasoning ofothers.): Students must identify which additional piece of information,if any, is needed to construct an argument that establishes the similarityof the given triangles.

Option A: This is the correct answer. Two pairs of corresponding congruentangles are sufficient for proving two triangles are similar. Since the sum of the

interior angles of all triangles is 180, the third corresponding pair of angles ofthe two triangles must be the same measure.

Option B: The student incorrectly assumes that information about thecorresponding sides of two triangles is necessary for proving similarity.

Option C: The student incorrectly assumes that three pairs of correspondingcongruent angles are necessary to prove two triangles are similar, which is nottrue because if two angles are known, then the third can be determined.

Option D: The student incorrectly uses the angle-side-angle (ASA) criterion forcongruence instead of AA criterion for similarity.


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Domain: CCircles

Cluster: AUnderstand and apply theorems about circles

Standard: 2Identify and describe relationships amonginscribed angles, radii, and chords. Include the

relationship between central, inscribed, andcircumscribed angles; inscribed angles on adiameter are right angles; the radius of a circleis perpendicular to the tangent where the radiusintersects the circle.





Use the figure to answer the question.

T

S Q

R

Triangle STR is drawn such that segment ST is tangent tocircle Q at point T, and segment SR is tangent to circle Q atpoint R. If given any triangle STR with these conditions, whichstatement must be true?

A. Side TR could pass through point Q.

B. Angle S is always smaller than angles T and R.

C. Triangle STR is always an isosceles triangle.

D. Triangle STR can never be a right triangle.

*correct answer


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This item requires students to identify properties of a triangle constructed usinga chord of a circle and two line segments tangent to the same circle. Specifically,students must recognize that tangent segments to a circle from the sameexternal point are congruent.


Math Practice 1 (Make sense of problems and persevere in solving them.):

Students must analyze the given information in the item and the constraintsof the details provided, and understand the goals of the problem.

Math Practice 3 (Construct viable arguments and critique the reasoning ofothers.): Students must determine which statement may be supported byproperties that exist based on the construction of the given triangle.

Option C: This is the correct answer. Tangent segments drawn from the samefixed point outside of the circle to points of tangency on the circumference ofthe circle are congruent. Therefore, the triangle is isosceles.

Option A: The student incorrectly assumes that a triangle could be created if

side TR passed through the center of the circle; however, if this were possible,then the lines tangent to the circle at T and R would be perpendicular tosegment TR. Thus, a triangle could not be created.

Option B: The student incorrectly assumes that angle S must always be smallerthan angle T and angle R; however, as points T and R move closer to each otheron the circumference of the circle, point S moves closer to the circle. Thiseventually allows angle S to grow larger than angle T and angle R.

Option D: The student incorrectly assumes that triangle STR cannot be aright triangle. It is true that angles T and R can never be right angles, but it ispossible for angle S to be equal to 90.


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Domain: GPEExpressing Geometric Properties withEquations

Cluster: ATranslate between the geometric description andthe equation for a conic section

Standard: 1Derive the equation of a circle of given center andradius using the Pythagorean Theorem; completethe square to find the center and radius of acircle given by an equation.

Math Practices: 7. Look for and make use of structure.

Calculator: Calculator not allowed


A circle has this equation.

x 2 + y 2 + 4x 10y = 7

What are the center and radius of the circle?

A. center: (2, 5)radius: 6

B. center: ( 2, 5)radius: 6

C. center: (2, 5)radius: 36

D. center: ( 2, 5)radius: 36

*correct answer

This item requires students to use the method of completing the square inorder to determine the center and radius of a circle given the equation.


Math Practice 7 (Look for and make use of structure.): Students mustcomplete the square for the equation to create an equivalent equation of theform (x h )2 + (y k )2 = r2. Then, they must use this structure to determinethat the center is (h, k ) and the radius is r.


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Option B: This is the correct answer. Completing the square in the originalequation results in (x + 2)2 + (y 5)2 = 62. Therefore, the center of the circle isthe point (2, 5) and the radius is 6 units.

GivenEquation

x 2 + y 2 + 4x 10y = 7

Step 1 x 2 + 4x + y 2 10y = 7 Rearrange x-termstogether and y-termstogether.

Step 2

(x 2 + 4x + ) + (y 2 10y + ) = 7+ 4

+4 +25 +25

(x2

+ 4x + 4) + (y2

10y + 25) = 36

Complete the squarefor the expressionx 2 + 4x by adding 4 toboth sides because4 2

2 = 4.

Complete the square

for the expressiony 210y by adding 25to both sides because

10 2 = 25.

2

Step 3 (x + 2)2 + (y 5)2 = 36Convert to factoredform.

Step 4

(x h )2 + (y k )2 = r2

(x (2))2 + (y 5)2 = 62h = 2 k = 5 r= 6

Find center at (h, k ).

Step 5

-

36 = 6 Find the radius.

Option A: This response results from a sign error when finding the center ofthe circle in step 4.

Option C: This response is the result of a sign error when finding the centerof the circle in step 4, and not taking the square root of the constant term todetermine the radius in step 5.

Option D: This response results from not taking the square root of the constantterm to determine the radius in step 5.


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Domain: GMDGeometric Measurement and Dimension

Cluster: AExplain volume formulas and use them to solveproblems

Standard: 3Use volume formulas for cylinders, pyramids,

cones, and spheres to solve problems.

Math Practices: 4. Model with mathematics.



Use the pyramid to answer the question.

8 in.

6 in.

This right pentagonal pyramid has a height of 8 inches and abase area of 61.94 square inches. To the nearest hundredth,what is the volume of the pyramid?

A. 80.00 cubic inches

B. 165.17 cubic inches

C. 240.00 cubic inches

D. 495.52 cubic inches

*correct answer


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This item requires students to calculate the volume of a right pentagonalpyramid given the area of the base.


Math Practice 4 (Model with mathematics): Students must identify thecorrect formula to calculate volume of a pentagonal pyramid. They must

apply the given information.

Option B: This is the correct answer. The formula for the volume of a

pyramid is V =1

Bh, where B is the area of the base and h is the height.3

For this pyramid, V =1

(61.94)(8) = 165.17 cubic inches.3

Option A: This response is the result of multiplying1

by the product of the3

height, the given side length, and the number of sides of the base.

Option C: This response results from computing the product of the height,

the given side length, and the number of sides of the base.

Option D: This answer results from failing to multiply1

by the product of the3

area of the base and the height.


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Domain: MGModeling with Geometry

Cluster: AApply geometric concepts in modeling situations

Standard: 2Apply concepts of density based on area andvolume in modeling situations (e.g., persons per

square mile, BTUs per cubic foot).


2. Reason abstractly and quantitatively.4. Model with mathematics.



Use the diagram to answer the question.

9 yards

12 yards

5 yardsAviary

An aviary is an enclosure for keeping birds. There are 134 birdsin the aviary shown in the diagram.

What is the number of birds per cubic yard for this aviary?Round your answer to the nearest hundredth.

A. 0.19 birds per cubic yard

B. 0.25 birds per cubic yard

C. 1.24 birds per cubic yard

D. 4.03 birds per cubic yard

*correct answer


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This item requires students to compute the volume of a rectangular prism.Then, students must use this value to determine the density of birds in anaviary.

This item is linked to three of the Math Practices.

Math Practice 1 (Make sense of problems and persevere in solving them.):Students must make sense of the given quantities and recognize that they

must first compute the volume to then find the quotient of the number ofbirds and the total volume of the aviary.

Math Practice 2 (Reason abstractly and quantitatively.): Students mustdecontextualize the given information and represent it in a logicalmathematical format. Then, they must manipulate this representationand contextualize their results to answer the question.

Math Practice 4 (Model with mathematics.): Students must model the givenreal-world situation using the equation for volume and the process todetermine density.

Option B: This is the correct answer. The volume of the aviary is given by theproduct of 5 yards, 12 yards, and 9 yards. The density of birds per cubic yardis given by the quotient of the number of birds and the volume of the aviary.

Option A: This response is the result of incorrectly using a sum of the givendimensions instead of calculating the volume of the aviary.

Option C: This answer results from incorrectly computing the number of birdsper square yard of floor space.

Option D: This response is the result of calculating the number of cubic yardsper bird rather than the number of birds per cubic yard.


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Domain: MGModeling with Geometry (MG)

Cluster: AApply geometric concepts in modeling situations

Standard: 3Apply geometric methods to solve designproblems (e.g., designing an object or structure

to satisfy physical constraints or minimize cost;working with typographic grid systems based onratios).


2. Reason abstractly and quantitatively.4. Model with mathematics.




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x

fenced-inarea

house

Note: not drawn to scale

width

Beth is going to enclose a rectangular area in back of her house.The house wall will form one of the four sides of the fenced-in

area, so Beth will only need to construct three sides of fencing.

Beth has 48 feet of fencing. She wants to enclose themaximum possible area.

What amount of fence should Beth use for the side labeled x?

A. 12 feet

B. 16 feet

C. 24 feet

D. 32 feet

*correct answer

This item requires students to solve an optimization problem involving area andperimeter. The most efficient pathway to solve the problem is to write equationsto represent the area and perimeter. Then, students may solve for one variable

in the linear perimeter formula and substitute this expression into the areaformula. This will lead to a quadratic equation representing the area in terms ofthe length of one of the sides. Finding the maximum point of this parabola willdetermine the side length that will maximize the area.

This item is linked to three of the Math Practices.

Math Practice 1 (Make sense of problems and persevere in solving them.):Students must make sense of the given quantities and recognize that theymust compute both area and perimeter.


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Math Practice 2 (Reason abstractly and quantitatively.): Students mustdecontextualize the given information and represent it in a logicalmathematical format. Then, they must manipulate this representationand contextualize their results to answer the question.

Math Practice 4 (Model with mathematics.): Students must model the givenreal-world situation with equations for the area and perimeter.

Option C: This is the correct answer.

Step 1x + 2w = 48, where w represents the widthA = x w

Determine necessaryequations to use.

Step 248 x

w =2

Solve perimeter formulafor w.

Step 348 x

A = x 2

Replace w in the Areaformula with equivalentexpression from step 2.

Step 4 A =

1

x2

+ 24x

2 Simplify the expression.

Step 5x =

b=

24= 24

2a2

1

2

Find the x-value of thevertex of the parabola,which represents wherethe area is maximized.

Therefore, the area is maximized when x = 24 feet.

Option A: This answer is a result of finding the width, w, for the maximized area.

Step 1x + 2w = 48, where w represents the widthA = x w

Determine necessaryequations to use.

Step 2 x = 48 2wSolve perimeter formulafor x.

Step 3 A = (48 2w)wReplace x withequivalent expressionfrom Step 2.

Step 4 A = 48w 2w2 Simplify expression.

Step 5 w = b

=48

=122a 2(2)

Find the width.

(incorrect to stop at

this point)Option B: The student incorrectly assumes that a square will maximize thearea of the fenced-in area without doing any calculations. This is true forsituations where all four sides of the enclosed area are being used.

Option D: The student incorrectly assumes that the largest given value for xwill result in the maximum area without doing any calculations.


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Constructed-Response Item

This section presents a constructed-response item and samples of studentresponses that received scores of 4, 3, 2, 1, 1 for minimal understanding, and 0.This section also includes information used to score this constructed-response

item: an exemplary response, an explanation of how points are assigned, and aspecific scoring rubric. In addition to the online resources available for all testquestions, students have access to the Geometry Typing Help (pages 44 and 45),which describes how to enter special characters, symbols, and formatting intotyped responses.

Constructed-Response Item



Cluster: CDefine trigonometric ratios and solve problemsinvolving right triangles

Standard: 8Use trigonometric ratios and the PythagoreanTheorem to solve right triangles in applied problems.


2. Reason abstractly and quantitatively.3. Construct viable arguments and critique the

reasoning of others.4. Model with mathematics.

6. Attend to precision.


Tool Restrictions: Protractor not allowedRuler not allowed


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(student enters response in text box)V


top of stairs

5 in.

12 in. 5 in.

12 in. 5 in.

Note: Not to scale

base of bottom stair

Leah needs to add a wheelchair ramp over her stairs. The rampwill start at the top of the stairs. Each stair makes a right anglewith each riser.

Part A

The ramp must have a maximum slope of1

. To the nearest12

hundredth of a foot, what is the shortest length of ramp that Leahcan build and not exceed the maximum slope? Show your workor explain your answer.

Length of ramp:

Work/Explanation:

(student enters response in text box)

V

Part BLeah decides to build a ramp that starts at the top of the stairsand ends 18 feet from the base of the bottom stair. To the nearesthundredth of a foot, what is the length of the ramp?

(student enters response in text box)V


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Part CTo the nearest tenth of a degree, what is the measure of the anglecreated by the ground and the ramp that Leah builds in part B?

(student enters response in text box)

V

This item requires students to reason about how slope, angle measures,trigonometric ratios, and the Pythagorean Theorem relate mathematically.Students must determine a horizontal measure for a leg of a right triangle

that results in a hypotenuse slope of no more than1

. Students then use the12

Pythagorean Theorem to find the measure of the hypotenuse in two differentright triangles. Finally, students must use a trigonometric ratio to determinethe measure of an angle in a right triangle.

This item is linked to five of the Math Practices.

Math Practice 1 (Make sense of problems and persevere in solving them.):Students must make sense of given information to determine how tosolve each part of the problem. The problem calls on multiple banks ofinformation and requires perseverance to solve each part.

Math Practice 2 (Reason abstractly and quantitatively.): Students mustdecontextualize given measurements to manipulate the numbers usingequations. Then, they must contextualize to determine the reasonablenessof their answer within the given context.

Math Practice 3 (Construct viable arguments and critique the reasoning ofothers.): Students must construct an argument to justify their reasoningbehind how they calculated the minimum length of ramp that will satisfythe given conditions in part A.

Math Practice 4 (Model with mathematics.): Students use the PythagoreanTheorem to model the relationship between the legs and hypotenuse ofthe right triangle used to calculate the length of the ramp. Students usetrigonometric ratios to model the relationship between side lengths andangle measures of the right triangle used to calculate the measure of theangle created by the ground and the ramp.

Math Practice 6 (Attend to precision.): Students must accurately convertbetween measurement units, specify the correct units of measure in theirfinal responses, and use proper rounding techniques.


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Scoring Information

Scoring Rubric

Score Description

4 The students response earns 4 points.



11 point or minimal understanding of using PythagoreanTheorem and trigonometric ratios.

0The students response is incorrect, irrelevant, too brief toevaluate, or blank.

Exemplary Response

Part ALength of ramp:15.05 feet

Work/Explanation:First find length of base of triangle (x)

1.25ft/x = 1/12

1.25*12 = 1*x

x = 15

Then find hypotenuse/ramp (c)

1.25^2 + 15^2 = c^2

226.5625 = c^2

c =* 15.05

Part B

20.04 feet

Part C

3.6 degrees

Points Assigned

Part A: 1 point for correct lengthPart A: 1 point for complete and accurate work or explanation

Part B: 1 point for correct length

Part C: 1 point for correct degree measure

Note:The student may receive credit for part C for correctly finding the anglemeasure based on an incorrect value from part B.


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Sample Student Responses

Score Point 4

The following authentic student responses show the work of two students whoeach earned a score of 4. A score of 4 is received when a student completes allrequired components of the task and communicates his or her ideas effectively.

The response should demonstrate in-depth understanding of the contentobjectives, and all required components of the task should be complete.

Score Point 4, Student 1

Part A15.05

To find the shortest length of the ramp, I first set the stairs up as a right trianlge.The height of the triangle(base to top of stairs) was 15 inches so I used the maximumslope to set it up a proportion to find the length of the bottom of the triangle. I found

the length would be 180 inches. I then used the pythagorean theorem to find thehypotenuse(length of the ramp) of the triangle. This answer was about 180.6324inches. However, the length of the ramp length needed to be in feet so I convertedit from inches to feet and got the shortest ramp possible would be 15.05 feet.

Part B20.04

Part C3.6 degrees

This student response is correct and well-reasoned. The student calculatesthe correct length and provides a correct and complete explanation of his or herwork for part A. The student also provides the correct length of the ramp inpart B and the correct angle measure in part C.


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Part A15.05 feet

1/12 = 1.25/15(1.25 x 1.25) + (15 x 15) = 226.56sqrt(226.56 = 15.052 feet

Part B20.04 feet

Part C3.6

This response receives full credit. The student calculates the correct lengthand provides a correct and complete explanation of his or her work for part A.

The student correctly calculates the length of the ramp in part B and thecorrect angle measure in part C.


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Score Point 3

The following authentic student responses show the work of two students whoeach earned a score of 3 for their responses. A score of 3 is received when astudent completes three of the four required components of the task. There maybe simple errors in calculations or some confusion with communicating his orher ideas effectively.


Part A15.05 feet

The shortest length of ramp that Leah can build and not exceed the maximum slopeis 15.05 feet. In order to get the length of the ramp, I first used the slope (1/12) asa ratio to find the length of the stairs. By multiplying the total heights of the risers,15 inches, by 12, I found that the total length of the stairs would be 180 inches.By using the Pythagorean Theorum to find the length of the hypotenuse, I found that

15^2 + 180^2 =* 180.62^2. I then divided the hypotenuse by 12 to get the length ofthe ramp in feet. 180.62 / 12 = 15.05 feet.

Part B19.04 feet

Part C3.8 degrees

This student calculates the correct length and provides a correct and

complete explanation of his or her work for part A. The student does notprovide the correct length of the ramp in part B; however, the studentsresponse in part C is correct based on this incorrect value in part B.


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Part A12.04 feet

I added the rise of each step to find the length of the short leg. For each inch of theshort leg I need 12 inches on the long leg. Then I used the pythagorean theorum tofind the hypotenuse to find the length of the ramp. Last, I converted it to feet.

Part B20.04 feet

Part C3.6

In this response, the student provides an incorrect length of the ramp inpart A. However, the student received partial credit of 1 point for providing a

correct and complete explanation of how to determine the length of the ramp.The student correctly calculates the length of the ramp in part B and thecorrect angle measure in part C.


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Score Point 2

The following authentic student responses show the work of two students whoeach earned a score of 2 for their responses. A score of 2 is received when astudent completes two of the four required components of the task. There maybe simple errors in calculations, one or two missing responses, or unclear orincorrect communications of his or her ideas.


Part A15.05

I used the pythagorem thereom to get find that , with a slope of 1/12, each in downwould take 12.04 inches. With this I could find the length of the ramp by multiplyingit by the height,15 inches. Which came out to be 180.6 inches. Next, I needed toconvert the answer to feet, which gave me 15.05 feet.

Part B18.04

Part C6.90%

This student provides the correct length of the ramp in part A and correctlyreasons that each vertical rise of 1 inch will result in a hypotenuse length of12.04 inches and multiplies that length by the height. The student does notprovide the correct length of the ramp in part B or the correct angle measure

in part C.


Part A15.05 feet

I used a ratio and Pythagorean Theorem. Then I found how many feet.

Part B20.04 feet

Part C30 degrees

The student provides the correct length for the ramp; however, the explanationis incomplete. There is not enough detail to receive full credit for their reasoning.The student provides the correct length of the ramp in part B, but the anglemeasure provided in part C is incorrect.


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Score Point 1

The following authentic student responses show the work of two students whoeach earned a score of 1 for their responses, and two students who each earneda score of 1 for demonstrating minimal understanding. A score of 1 is receivedwhen a student correctly addresses one of the four required components, ordemonstrates at least minimal understanding of the key concepts.


Part A180.62

First I started out with what I know is going to be the height of the triangle we aregoing to make with the ramp and the floor. Since slope = rise/run I then figured outthat the base of our tiangle would have to be 180in for the ratio to match 1/12. Afterthat I simply used the pathagoream therom to figue out the hypotenuse or in thiscase, our ramp.

Part B216.52

Part C4

This student provides the correct length of the ramp in part A, but expressesthe measurement in the wrong unit. The item specifically asks for the lengthof the ramp in feet; however, the reasoning provided is correct and complete,

so the student earned 1 point. The student does not provide the correct lengthof the ramp in part B or the correct angle measure in part C.


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Part A15 feet

To the top of the stairs is 15 inches. So every inch you go up you have to go over onefoot. Which would make the ramp 15 feet long.

Part B20 feet

Part C3.6 degrees

The student provides an incorrect ramp length. While the student correctlyexplains how to determine the measure of the long leg of the right triangle,the response is incomplete because the student does not provide a method

for calculating the length of the hypotenuse. The student does not provide thecorrect length of the ramp in part B. However, the students response in part Cis correct based on this incorrect value in part B.

Score Point 1 (minimal understanding), Student 3

Part A180.62

Because the ramp cannot exeed a slope of (1/12), the angle of the ramp cannotexeed 4.76 degrees. Using trigonometry, you can find the length of the ramp.

Part B44.6

Part C19.65

This student provides the correct length of the ramp in part A in inches,but does not provide a unit of measure. The explanation is also incompleteas it does not provide a method for computing the hypotenuse of the righttriangle. The student does not provide the correct ramp length in part B orthe correct angle measure in part C. This response does not earn any points;however, one point is awarded for minimal understanding of the PythagoreanTheorem for providing the correct length of the ramp in part A using the wrongmeasurement unit.


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Score Point 1 (minimal understanding), Student 4

Part A39 in.

A squared plus B squared equals C squared. 5 is A. 12 is B. 5 squared equals 25.12 squared equals 144. 25 plus 144 equals 169. The square root of 169 is 13.There are three 5 inch drops. Multiply 13 in by 3 and that will equal 39 in.

Part B240.46 in.

Part C3.5

This student does not provide the correct ramp length in part A. The explanationdoes not take into account the slope of the ramp. While the Pythagorean Theorem

is used correctly, it is applied using the wrong numbers. The student providesthe correct ramp length in part B, but uses incorrect units of measure.The student provides an angle measure in part C that is a result of truncatingthe correct response instead of rounding to the nearest tenth. This responsedoes not earn any points; however, one point is awarded for minimalunderstanding of using trigonometric ratios to solve problems for part C.One point for minimal understanding could also be awarded for part B;however, only one minimal understanding point may be awarded per item.


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Score Point 0

The following samples show the work of two students who each earned a scoreof 0. A score of 0 is received when a student response is incorrect, irrelevant,too brief to evaluate, or blank.


Part A12/13

12^2 + 5^2= 169 sqrt(169)= 13 rise/run 24/26 = 12/13

Part B51 feet

Part C90 degrees

This student does not provide the correct length of the ramp in part A.The reasoning provided uses the Pythagorean Theorem, but it is applied usingthe wrong numbers. There is no mention of the slope of the ramp. The studentdoes not provide the correct ramp length in part B or the correct angle measurein part C.


Part A0.0ft

If you divide 12 into 1 it comes out to be 0.083. round that to the nearest hundredth ofa foot it comes out to be 0.1ft., anything less then 0.ft is 0.0ft.

Part B1.5ft

Part C2ft

This student does not provide the correct length of the ramp in part A.

The reasoning provides no useful information for calculating the length ofthe ramp. The student does not provide the correct ramp length in part B.The answer provided for the missing angle in part C is incorrect in value andis expressed in the wrong unit of measure.


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Geometry Typing Help

Keystrokes for Special Symbols

1. If the ResponseIncludes:

2. Type This Instead: 3. Example:

multiplication symbol

x letter

x OR*

asterisk (SHIFT + 8)

3 x 4 = 12

3 * 4 = 12

division symbol

/forward slash

12 / 3 = 4

fraction or ratio

/forward slash

(12 7)/(3 1)Note: Parentheses

are required.

2 34

mixed number

space between whole

number and fraction;forward slash to separate

numerator anddenominator of fraction

2 3/4

32exponent

^caret (SHIFT + 6)

3^2 = 9

pi symbol(pi) Area = 9(pi)

square inches

greater than or equal to

>=greater than sign, followed

by equals sign

y >= 13

less than or equal to


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Keystrokes for Special Symbols (continued)

angle symbol

angle(spell out the word)

Angle PQR is a right angle.

ABCtriangle symbol

triangle(spell out the word)

Triangle ABC is a righttriangle.

m qperpendicular symbol

perpendicular(spell out the word)

Line m is perpendicular toline q.

m || q

parallel symbol

||two pipes (SHIFT +

backslash)OR parallel

(spell out the word)

m || qOR

Line m is parallel to line q.

ABC RSTcongruence symbol

congruent(spell out the word)

Triangle ABC is congruentto triangle RST.

STU VWX

similarity symbolsimilar

(spell out the word)Triangle STU is similar to

triangle VWX.

line segmentline segment

(spell out the words)Line segment AB bisects

line segment CD.

lineline

(spell out the word)Line AB is parallel to

line CD.

ray ray(spell out the word) Ray AB goes throughpoint P.


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Geometry Reference SheetUse the information below to answer

questions on the Geometry test.

Pythagorean Theorem

ca

a 2 + b 2 = c 2

b

Sphere

r

3.14

2Surface Area = 4r

Volume =4r 3

3

Cone

h s

Surface Area rs r 2

1 2

Volume = r hr 3

Cylinderr

h

Surface Area = 2r2+ 2rh

Volume = r 2h

Regular PyramidB = area of base

L = area of lateral surfacess h Surface Area = B + L

b Volume = 1 Bh3

Rectangular Solid

h Surface Area = 2wl + 2lh + 2wh

Volume = lwhw

l


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2 1 12

Circle

rArea = r

2

Circumference = 2r

Rectangle

w

l

Area = lw

Perimeter = 2l 2w

Trapezoid

b1

Area =1h(b + b )

h 2 1 2

b2

Triangle

h

b

Area =1bh

2

Parallelogram

h

b

Area = bh

Cartesian DistanceFormula

Point 1: (x1, y1 )

Point 2: (x , y )

d = (x

2 2

x )2 + ( y y )2


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